13
$\begingroup$

On a warm and uneventful summer day, John and Mary passed the time by playing games of Tic-Tac-Toe, taking sticks and drawing in the soil. John and Mary are both child prodigies, so within two rounds they solved the game and got bored. To decrease their boredom, Mary proposed a new and exciting variant of Tic-Tac-Toe.

In normal games of Tic-Tac-Toe, a 3x3 grid is drawn pre-game, and is then filled with the Xs and Os. In Mary's variant, the grid is only hypothetical, and it's position is not yet known. It is up to the players to assert it's exact location in making their moves.

Mary is an extremely confusing child, and she often scares her parents with how strange she is, so naturally we must walk through a sample game to understand what is going on in her head.


1st

The first move is made. Note that although a border is drawn, Mary's game takes place on an infinite plane without a border. The entire game takes place within this square for convenience.


2nd

The second move. An observant one would note that these first two moves do not actually change the course of the game.


3rd

Player One asserts a possible location of the 3x3 board based on the given moves, and follows up on his hypothesis with his third move.


4th

Player Two retorts, proclaiming a different possible position of the board, and follows up with the fourth move. With this move, the board is now uniquely defined with the given information. Play can now continue normally.


5th

A sequence of forced moves shows that Player Two's defiance completely backfires, facing a double threat, and she loses the game.


In Mary's Tic-Tac-Toe variation, what will be the outcome with optimal play? Does Player One have a forced win?

$\endgroup$
  • $\begingroup$ Is the plane the two play on of infinite size? And: Has the grid to be uniform, or does it just have to resemble the general shape, with arbitrarily wide/high columns and rows? Do all columns/rows have to have the same width/height $\endgroup$ – Verzweifler Apr 24 '17 at 11:09
  • $\begingroup$ The plane is infinite and the grid is a uniform 3x3 square grid. $\endgroup$ – greenturtle3141 Apr 24 '17 at 11:28
  • $\begingroup$ Should symbols be in the center of their cells? $\endgroup$ – Alix Eisenhardt Jun 2 '17 at 10:15
  • $\begingroup$ @Alix Eisenhardt Yes. $\endgroup$ – greenturtle3141 Jun 2 '17 at 12:30
  • $\begingroup$ In that case, Curtis' answer is the solution. $\endgroup$ – Alix Eisenhardt Jun 2 '17 at 13:20
12
$\begingroup$

X plays, O plays, X plays as follows:

X O

X

We're now locked on this plane, albeit not uniquely (the O can be in the central or rightmost column). Either way, O is forced:

X O
O
X

And now X can uniquely define the board and force a double threat, and thus a win:

X O
O
X   X
$\endgroup$
  • $\begingroup$ I think this depends on whether or not the plane they played on is of infinite size - in step 2, "O" could be set below the 2nd "X", filling the row while blocking the double threat. Though this is only possible if columns may be of varying widths... $\endgroup$ – Verzweifler Apr 24 '17 at 11:10
  • 1
    $\begingroup$ @Verzweifler: I assumed from the diagrams at the top that squares had to be square; looking over it though, that's not explicitly stated anywhere... So, depends on the asker's intent, I guess. $\endgroup$ – BlueHairedMeerkat Apr 24 '17 at 11:17
  • 1
    $\begingroup$ Since OP just statet that the fields have to be uniform squares, I agree with your answer. Furthermore, since the playing area is infinite, player 1 could even treat player 2's first "O" always as put in "top middle". But as you wrote, he doesn't have to. $\endgroup$ – Verzweifler Apr 24 '17 at 11:40
  • $\begingroup$ @Verzweifler Unless you are required that your entries are in the middle of the square adding the second O above or below the first 2 is possible. It just forces the first X to be far to the right in the square, and the first O to be far to the left in it's square. $\endgroup$ – Taemyr Apr 24 '17 at 12:19
2
$\begingroup$

Yes, as long as I didn't interpret you wrong, I believe that player 1 has a forced win:

Diagram:

enter image description here

Diagram explantion and reasoning:

If The second player plays expecting the first player to go to the corners (The best strategy for winning) then the player should be trying for the center, if not they will loose as long as the other player dosn't make a mistake.

2.

On the top left Tic-Tac-Toe game the second player plays as if the 1st player played in the corner and plays in the center according to the way he or she sees it. This however is inefective because the third play can define the boundaries of the game.

3.

In the top right Tic-Tac-Toe game the 1st player can look at the 2nd player's move as dirrectly beside theirs and play to the corner farthest from their own. (This will make player 1 win.)

4.

The third bottom Tic-Tac-Toe game just shows that player 2 is now restricted and can no longer win any way

$\endgroup$
  • 1
    $\begingroup$ Why would player 2 place the second O there? Surely they would place it in the centre square, blocking the direct threat? $\endgroup$ – Fil-let's GoFundMonica Apr 24 '17 at 8:09
  • $\begingroup$ @Fillet same thing I was wondering myself. $\endgroup$ – The Great Duck Apr 24 '17 at 8:19
  • $\begingroup$ I think player 2 tried to place his second "O" between the two "X" of player 1 (a forced move, see image #2). Player 1 then placed his third mark such that he "rotates" the board to the position shown in image #3, making the diagonal into the new rightmost row. But I think that whether this "rotation" is possible or not depends on the size and exact location of drawn Shapes. $\endgroup$ – Verzweifler Apr 24 '17 at 9:03
  • $\begingroup$ @Verzweifler That makes sense, but I'm pretty sure the rotation isn't possible. If you overlay a chess board on image 2, then the Xs are on the black squares, and the first O is on a white square. The proposed block would also be on a black square. The planned rotation would leave only the black squares playable, and the first O is now in an "unplayable" gap, equally far from the new centre spot and new center-top spot. $\endgroup$ – Fil-let's GoFundMonica Apr 24 '17 at 9:41
  • $\begingroup$ This is not optimal play for O, they can just place their O directly between the two X's, and given that the first three moves have fixed the board this must be the centre. $\endgroup$ – BlueHairedMeerkat Apr 24 '17 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.